Quote Originally Posted by irving2008 View Post
You'll also need a bleed resistor across each capacitor to discharge it when the power is turned off (for safety). The energy stored is 0.5CV^2 joules. For each capacitor spec'd above, the energy is 0.5 * .005 * 70^2 = 6joules. A watt is a joule per second, so for a 5 second discharge you need to dissipate 6/5 = 1.2W so we're going to need 2W rated resistors. Assuming the resistor dissipates the same 1.2W when the power supply is turned on the value will be R = v^2/W = 70^2/1.2 = 4000ohm. I'd use 3900ohm (3k9) @ 2W wire ended and solder them directly to the ring tags on the capacitor: Buy Through Hole Fixed Resistors Carbon Resistor, 2W ,5%, 3k9 RS RS-Carbon-3k9-5%-2W online from RS for next day delivery. (pack of 10 @ 72p)
After a cup of tea and a bit of a read, I have a quick question in relation to 0.5CV^2. Where you give 0.5 * 0.005 * 70^2 = 6, I end up with 12.2. In fact using the time honoured technique of "randomly changing numbers until they worked", I only got 6j by changing the 70v to 16v?

Trying to follow your calculation and sticking with the 5 second discharge, I ended up with this:
0.5 * 0.005 * 70^2 = 12.25j
12.25 / 5 = 2.45W
70^2/2.45 = 2000ohm, so I would need a 2k ohm @ 3W resistor to discharge over 5 seconds.

I'm rather hoping you made a typo because the alternative is that I'm more stupid than I thought!